Answer to Question #276619 in Statistics and Probability for Yara

Question

Among drivers who have had a car crash in the last year, 180 were randomly selected and categorized by age, with the results listed in the table below.

Age. Under 25. 25-44. 45-64. Over 64

Drivers. 72 40 28 40

If all ages have the same crash rate, we would expect (because of the age distribution of licensed drivers) the given categories to have 16%, 44%, 27%, 13% of the subjects, respectively. At the 0.05 significance level, test the claim that the distribution of crashes conforms to the distribution of ages.

The test statistic is 𝜒2 =

The critical value is 𝜒2 =

 

Answer 1

The hypotheses tested are,

$H_0:$ The distribution of crashes conforms to the distribution of ages.

$Against$

$H_1:$ The distribution of crashes does not conform to the distribution of ages.

To perform this test we shall apply the chi-square goodness of fit test.

We determine the expected frequency for each cell using the formula below,

$E_i=n*p_i$ where $p_i=0.16,0.44,0.27,0.13$ for $i=1,2,3,4$ and $n=180$

The expected counts are,

$E_1=n*p_1=180*0.16=28.8$

$E_2=n*p_2=180*0.44=79.2$

$E_3=n*p_3=180*0.27=48.6$

$E_4=n*p_4=180*0.13=23.4$

The test statistic is given as,

$\chi^2_c=\displaystyle \sum^4_{i=1}(O_i-E_i)^2/E_i$

Now,

$\chi^2_c=(72-28.8)^2/28.8+(40-79.2)^2/79.2+(28-48.6)^2/48.6+(40-23.4)^2/23.4=104.7098$

$\chi^2_c$ is compared with the chi square table value at $\alpha=5\%$ with $(i-1)=(4-1)=3$, where $i$ is the age categories for a driver.

The table value is given as, $\chi^2_{\alpha, 3}=\chi^2_{0.05,3}=7.81473$ and the null hypothesis is rejected if $\chi^2_{c}\gt \chi^2_{0.05,3}$.

Since $\chi^2_{c}=104.7098\gt \chi^2_{0.05.3}=7.81473$, we reject the null hypothesis and conclude that there is no sufficient evidence to show that the  distribution of crashes conforms to the distribution of ages at 5% significance level.